Integer approximation of a segment

Let $OXY$ be a Cartesian coordinate system with an integer lattice whose unit squares are staggered. The integer approximation of the segment $AB$ is given using the cellular domain $\mathbb{S}_{AB}$ of (colored) cells, the interior of each of which.has a non-empty intersection with $AB$. If $P^{\pm}_{AB}$ — right and left closed half-planes defined by the line $l_{AB}$ by the point $A$ and $B$, then $\mathbb{S}^{\pm}_{AB}=\mathbb{S}_{AB}\cap P^{\pm}_{AB}$ — its right and left areas. (There are no integer points inside $\mathbb{S}_{AB}$.) Polyline $\mathrm{L}^{\pm}(A^{\pm},B^{\pm})$ from $\mathbb{S}^{\pm}_{AB}$ with ends $A^{\pm}$ and $B^{\pm}$ and whole vertices — right and left by (integer) approximations of the segment $AB$ — the ends are selected from the vertices of the extreme cells. If $l_{AB}$ is parallel to one of the coordinate axes, then we assume $\mathbb{ S}_{AB}=\varnothing$ and then approximation of the segment $AB$ is minimum segment with integer ends containing $AB$. Such approximations are constructed using the algorithm ‘pulling noses’, which is a geometric interpretation of the chain fraction of the angular coefficient of the straight line $l_{AB}$. Based on this construction method, an exact formula for calculating the number of integer points inside an arbitrary triangle is obtained, and the problem of S.V. Konyagin is partially solved. about chess coloring: If $\mathbf{U}(t)$ is the set of all colored cells from a triangle cut off by a straight line $f_{t}:y= -\alpha x+t$, $\alpha,\ t>0$, then the difference $u(t)$ between white and black cells from $\mathbf{U}(t)$ for every positive irrational $\alpha$ is bounded neither from below nor from above when $t\rightarrow\infty$. The solution is obtained for numbers of the form: $e^{\pm1}$, $\mathrm{tg}^{\pm1}$, $[a^{-}_{0}; a^{-}_{1},a^{-}_{2},\ldots]^{\pm1}$, $[a^{+}_{0};a^{+}_{1},a^{+}_{2}, \ldots]^{\pm1}$, $[a^{+}_{0};a^{-}_{1},a^{+}_{2}, \ldots]^{\pm1}$, where the superscript plus (minus) indicates on the parity (odd) of the element of the continued fraction defined by $\alpha$.
The method of constructing an approximation of the segment was used to solve the problem of chess coloring for the numbers $\frac{\sqrt{5}+1}{2}$, $[ a^{+}_{0};a^{+ }_{1},a^{+}_{2},\ldots]$, $a^{-}_{2n+1}$ and $[a^{-}_{0};a^{-}_{1},a^{-}_{ 2}, \ldots]$, if limited
\begin{equation*} \begin{array}{l} 2^{k-1}b_{3}b_{9} \cdots b_{6(k-1)+3} + \cdots + 2^{2}\sum^{k}_{i_{1}>i_{2}>i_{3}=1} b_{6(k-i_{1})+3}b_{6(k-i_{2})+3} \\ b_{6(k-i_{3})+3} + 2\sum^{k}_{i_{1}>i_{2}=1} b_{6(k-i_{1})+3}b_{6(k-i_{2})+3} + \sum^{k}_{i=1}b_{6(k-i_{1})+3} + 1, \end{array} \end{equation*}
for some $b_{n}=\left\lfloor\frac{a^{-}_{n}-1}{2}\right \rfloor$ representing the whole part of $\frac{a^{-}_{n}-1}{2}$. So for $b_{n}=0$ the chain fraction is $[a^{-}_{0};a^{-}_{1},a^{-}_{2}, \ldots]=\frac{\sqrt{5}+1}{2}$.